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Commit e6fb81d

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Add knapsack problem (TheAlgorithms#2330)
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package DynamicProgramming;
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/* A Naive recursive implementation
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of 0-1 Knapsack problem */
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public class BruteForceKnapsack {
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// A utility function that returns
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// maximum of two integers
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static int max(int a, int b) {
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return (a > b) ? a : b;
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}
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// Returns the maximum value that
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// can be put in a knapsack of
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// capacity W
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static int knapSack(int W, int wt[], int val[], int n) {
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// Base Case
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if (n == 0 || W == 0)
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return 0;
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// If weight of the nth item is
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// more than Knapsack capacity W,
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// then this item cannot be included
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// in the optimal solution
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if (wt[n - 1] > W)
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return knapSack(W, wt, val, n - 1);
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// Return the maximum of two cases:
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// (1) nth item included
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// (2) not included
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else
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return max(val[n - 1] + knapSack(W - wt[n - 1], wt, val, n - 1), knapSack(W, wt, val, n - 1));
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}
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// Driver code
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public static void main(String args[]) {
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int val[] = new int[] { 60, 100, 120 };
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int wt[] = new int[] { 10, 20, 30 };
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int W = 50;
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int n = val.length;
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System.out.println(knapSack(W, wt, val, n));
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}
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}
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package DynamicProgramming;
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// A Dynamic Programming based solution
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// for 0-1 Knapsack problem
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public class DyanamicProgrammingKnapsack {
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static int max(int a, int b) {
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return (a > b) ? a : b;
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}
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// Returns the maximum value that can
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// be put in a knapsack of capacity W
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static int knapSack(int W, int wt[], int val[], int n) {
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int i, w;
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int K[][] = new int[n + 1][W + 1];
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// Build table K[][] in bottom up manner
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for (i = 0; i <= n; i++) {
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for (w = 0; w <= W; w++) {
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if (i == 0 || w == 0)
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K[i][w] = 0;
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else if (wt[i - 1] <= w)
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K[i][w] = max(val[i - 1] + K[i - 1][w - wt[i - 1]], K[i - 1][w]);
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else
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K[i][w] = K[i - 1][w];
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}
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}
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return K[n][W];
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}
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// Driver code
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public static void main(String args[]) {
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int val[] = new int[] { 60, 100, 120 };
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int wt[] = new int[] { 10, 20, 30 };
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int W = 50;
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int n = val.length;
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System.out.println(knapSack(W, wt, val, n));
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}
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}
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package DynamicProgramming;
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// Here is the top-down approach of
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// dynamic programming
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public class MemoizationTechniqueKnapsack {
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//A utility function that returns
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//maximum of two integers
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static int max(int a, int b) {
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return (a > b) ? a : b;
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}
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//Returns the value of maximum profit
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static int knapSackRec(int W, int wt[], int val[], int n, int[][] dp) {
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// Base condition
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if (n == 0 || W == 0)
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return 0;
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if (dp[n][W] != -1)
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return dp[n][W];
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if (wt[n - 1] > W)
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// Store the value of function call
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// stack in table before return
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return dp[n][W] = knapSackRec(W, wt, val, n - 1, dp);
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else
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// Return value of table after storing
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return dp[n][W] = max((val[n - 1] + knapSackRec(W - wt[n - 1], wt, val, n - 1, dp)),
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knapSackRec(W, wt, val, n - 1, dp));
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}
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static int knapSack(int W, int wt[], int val[], int N) {
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// Declare the table dynamically
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int dp[][] = new int[N + 1][W + 1];
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// Loop to initially filled the
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// table with -1
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for (int i = 0; i < N + 1; i++)
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for (int j = 0; j < W + 1; j++)
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dp[i][j] = -1;
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return knapSackRec(W, wt, val, N, dp);
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}
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//Driver Code
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public static void main(String[] args) {
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int val[] = { 60, 100, 120 };
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int wt[] = { 10, 20, 30 };
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int W = 50;
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int N = val.length;
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System.out.println(knapSack(W, wt, val, N));
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}
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}

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